Q. The solution of the differential equation $\left( 1 + y^{2} \right)\tan^{- 1}x\ dx + y\left( 1 + x^{2} \right)dy = 0$ is
A) $\log{\left( \frac{\tan^{- 1}x}{x} \right) + y\left( 1 + x^{2} \right) = c}$
B) $\log{\left( 1 + y^{2} \right) + \left( \tan^{- 1}x \right)^{2} = c}\$
C) $\log{(1 + x^{2}}) + \log{(\tan^{- 1}y}) + c$
D) $\left( \tan^{- 1}x \right)\left( 1 + y^{2} \right) + c = 0$
Explanation
(b)
Given, $\frac{\tan^{- 1}x}{1 + x^{2}}dx + \frac{y}{1 + y^{2}}dy = 0$
$\Rightarrow$ $\frac{\left( \tan^{- 1}y \right)^{2}}{2} + \frac{1}{2}\log{(1 + y^{2})} = \frac{c}{2}$ [integrating]
$\Rightarrow$ $\left( \tan^{- 1}x \right)^{2} + \log{(1 + y^{2})} = c$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.